Question

Photoelectric emission takes place from a certain metal at threshold frequency \( \nu \). If the radiation of frequency \( 4\nu \) is incident on the metal plate, the maximum velocity of the emitted photoelectrons will be (\( m = \) mass of photoelectron, \( h = \) Planck's constant)}

• A. \( \sqrt{\frac{6h\nu}{m}} \)
• B. \( \sqrt{\frac{3h\nu}{m}} \)
• C. \( \sqrt{\frac{h\nu}{m}} \)
• D. \( \sqrt{\frac{5h\nu}{m}} \)

Hint:

$K.E. = h(\text{Incident Frequency} - \text{Threshold Frequency})$.

Answer 1

The Correct Option is A

Step 1: Einstein's Equation
$h\nu' = \phi + K.E._{max}$.
Threshold frequency is $\nu$, so Work function $\phi = h\nu$.
Step 2: Substitute Values
Incident frequency $\nu' = 4\nu$.
$h(4\nu) = h\nu + \frac{1}{2}mv^2$.
Step 3: Solve for v
$3h\nu = \frac{1}{2}mv^2 \implies v^2 = \frac{6h\nu}{m} \implies v = \sqrt{\frac{6h\nu}{m}}$.
Final Answer: (A)